I'm reading a book that states the following:
Suppose we have a continuously differentible function $f(t)$. Around each $t_0$ we can expand $f(t)$, $$f(t) = f(t_0) + f'(t_0)(t-t_0) + o(t-t_0)$$
Shouldn't the last term read $o\big((t-t_0)^2\big)$? Initially I thought it was a mistake, but they kept writing it, so now I think there's something I'm missing.
On
No. You're probably confusing $\;o\bigl((t-t_0)^2\bigr)$, which means that $\lim_{t\to t_0}\dfrac{o\bigl((t-t_0)^2\bigr)}{(t-t_0)^2}=0$ (Landau's notation) and $\;O\bigl((t-t_0)^2\bigr)$, which means that $\dfrac{O\bigl((t-t_0)^2\bigr)}{(t-t_0)^2}$ is bounded in a small neighbourhood of $t_0$ ((Bachmann's notation).
Taylor's formula with $o\bigl(((t-t_0)^n\bigr)$ is known as Taylor-Young's formula.
On
No. $o((t-t_0)^2)$ is impossible; consider $f(t)=t^2$, $t_0=0$.
The last term could be improved to $O((t-t_0)^2)$ under stronger hypotheses. But given just that $f$ is continuously differentiable $o(t-t_0)$ is exactly right; if you sort through the definitions you see that $$f(t)=f(t_0)+A(t-t_0)+o(t-t_0)$$is precisely equivalent to $$\lim_{t\to t_0}\frac{f(t)-t(t_0)}{t-t_0}=A,$$or $$f'(t_0)=A.$$
No. $f(x)=o(g(x))$ as $x \to x_0$ if $f(x)/g(x) \to 0$ as $x \to x_0$. Indeed, the equation the book gives is equivalent to the definition of the derivative: $f$ is differentiable at $t_0$ if and only if there is a number $f'(t_0)$ so that $$ \frac{f(t)-f(t_0)}{t-t_0} - f'(t_0) \to 0 $$ as $t \to t_0$, which is the same as saying that $f(t) = f(t_0) + f'(t_0)(t-t_0) + o(t-t_0)$.
On the other hand, $f(x)=O(g(x))$ as $x \to x_0$ just means that $ \lvert f(x) \rvert \leq C\lvert g(x) \rvert$ for some fixed $C$ for $x$ sufficiently close to $x_0$.
The function $f(x)=x^{4/3}$ provides an example that shows that continuously differentiable does not give you $O(x^2)$: we have $f(0)=f'(0)=0$, but it is certainly not true that $x^{4/3} \leq C x^2$ for some fixed $C$ as $x \to 0$ since $ x^{4/3-2} = x^{-2/3} \to \infty$ as $x \to 0$.